Integrand size = 30, antiderivative size = 125 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{45 d e^4 \sqrt {e \sec (c+d x)}}-\frac {8 i a (a+i a \tan (c+d x))^{3/2}}{45 d e^2 (e \sec (c+d x))^{5/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}} \] Output:
-16/45*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/e^4/(e*sec(d*x+c))^(1/2)-8/45*I*a* (a+I*a*tan(d*x+c))^(3/2)/d/e^2/(e*sec(d*x+c))^(5/2)-2/9*I*(a+I*a*tan(d*x+c ))^(5/2)/d/(e*sec(d*x+c))^(9/2)
Time = 1.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {a^2 (9+25 \cos (2 (c+d x))-20 i \sin (2 (c+d x))) (-i \cos (2 (c+2 d x))+\sin (2 (c+2 d x))) \sqrt {a+i a \tan (c+d x)}}{45 d e^4 \sqrt {e \sec (c+d x)} (\cos (d x)+i \sin (d x))^2} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(9/2),x]
Output:
(a^2*(9 + 25*Cos[2*(c + d*x)] - (20*I)*Sin[2*(c + d*x)])*((-I)*Cos[2*(c + 2*d*x)] + Sin[2*(c + 2*d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(45*d*e^4*Sqrt[e *Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])^2)
Time = 0.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3978, 3042, 3978, 3042, 3969}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}}dx\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle \frac {4 a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{(e \sec (c+d x))^{5/2}}dx}{9 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{(e \sec (c+d x))^{5/2}}dx}{9 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle \frac {4 a \left (\frac {2 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}}\right )}{9 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a \left (\frac {2 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}}\right )}{9 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {4 a \left (-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}}\right )}{9 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(9/2),x]
Output:
(((-2*I)/9)*(a + I*a*Tan[c + d*x])^(5/2))/(d*(e*Sec[c + d*x])^(9/2)) + (4* a*((((-4*I)/5)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*Sqrt[e*Sec[c + d*x]]) - (((2*I)/5)*(a + I*a*Tan[c + d*x])^(3/2))/(d*(e*Sec[c + d*x])^(5/2))))/(9 *e^2)
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ [Simplify[m + n], 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/( a*f*m)), x] + Simp[a*((m + n)/(m*d^2)) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b ^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
Time = 6.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (5 \,{\mathrm e}^{4 i \left (d x +c \right )}+18 \,{\mathrm e}^{2 i \left (d x +c \right )}+45\right )}{90 e^{4} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(89\) |
default | \(-\frac {2 \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (10 i \cos \left (d x +c \right )^{4}-10 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}-i \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right ) \sin \left (d x +c \right )+8 i\right )}{45 d \sqrt {e \sec \left (d x +c \right )}\, e^{4}}\) | \(92\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/90*I*a^2/e^4/(e*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^(1/2)*(a*exp(2*I*( d*x+c))/(exp(2*I*(d*x+c))+1))^(1/2)*(5*exp(4*I*(d*x+c))+18*exp(2*I*(d*x+c) )+45)/d
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {{\left (-5 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 23 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 63 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 45 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{90 \, d e^{5}} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x, algorithm="fric as")
Output:
1/90*(-5*I*a^2*e^(6*I*d*x + 6*I*c) - 23*I*a^2*e^(4*I*d*x + 4*I*c) - 63*I*a ^2*e^(2*I*d*x + 2*I*c) - 45*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt( e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^5)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(9/2),x)
Output:
Timed out
Time = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {{\left (-5 i \, a^{2} \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) - 18 i \, a^{2} \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 45 i \, a^{2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{2} \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 \, a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 45 \, a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{90 \, d e^{\frac {9}{2}}} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x, algorithm="maxi ma")
Output:
1/90*(-5*I*a^2*cos(9/2*d*x + 9/2*c) - 18*I*a^2*cos(5/2*d*x + 5/2*c) - 45*I *a^2*cos(1/2*d*x + 1/2*c) + 5*a^2*sin(9/2*d*x + 9/2*c) + 18*a^2*sin(5/2*d* x + 5/2*c) + 45*a^2*sin(1/2*d*x + 1/2*c))*sqrt(a)/(d*e^(9/2))
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 2.82 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=-\frac {a^2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-18\,\sin \left (c+d\,x\right )-23\,\sin \left (3\,c+3\,d\,x\right )-5\,\sin \left (5\,c+5\,d\,x\right )+\cos \left (c+d\,x\right )\,108{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,23{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,5{}\mathrm {i}\right )}{180\,d\,e^5} \] Input:
int((a + a*tan(c + d*x)*1i)^(5/2)/(e/cos(c + d*x))^(9/2),x)
Output:
-(a^2*(e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(c + d*x)*108i - 18*sin(c + d*x) + cos(3*c + 3*d*x)*23i + cos(5*c + 5*d*x)*5i - 23*sin(3*c + 3*d*x) - 5*sin(5 *c + 5*d*x)))/(180*d*e^5)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{5}}d x \right )+2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{5}}d x \right ) i +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right )^{5}}d x \right )}{e^{5}} \] Input:
int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x)
Output:
(sqrt(e)*sqrt(a)*a**2*( - int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1) *tan(c + d*x)**2)/sec(c + d*x)**5,x) + 2*int((sqrt(sec(c + d*x))*sqrt(tan( c + d*x)*i + 1)*tan(c + d*x))/sec(c + d*x)**5,x)*i + int((sqrt(sec(c + d*x ))*sqrt(tan(c + d*x)*i + 1))/sec(c + d*x)**5,x)))/e**5